IConstant field in the Lagrangian

Hello! If you have a Lagrangian (say of a scalar field) depending only on the field and its first derivative and you want to calculate the ground state configuration, is it necessary a constant value? I read about Spontaneous symmetry breaking having this Lagrangian $$L= \frac{1}{2}(\partial_\mu \phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4$$ and they say that to get the lowest energy configuration you set ##\phi## to be a constant and work from there. In the case of a particle it makes sense that a motionless particle would have less energy that a moving one at a point. but is this obvious for fields? Can't a complicate configuration bring the energy lower than a constant value? Thank you!

You can probably answer your own question by computing the energy density of your scalar field.

Sorry I did a mistake, the term mass should be + not - (in order to get spontaneous symmetry breaking). So the energy density would be $$H= \frac{1}{2}(\partial_\mu \phi)^2-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$. Is it obvious that the field must be constant?

No, that is not the correct energy density. Are you familiar with how the energy-momentum tensor is derived from the Lagrangian?

I am not sure I am. But we were told that the hamiltonian density is ##H=T+V## which corresponds to the energy density (when you integrate it you get the actual hamiltonian which gives the energy for well behaved potentials). Did I miss-understand that?

I am not sure I am. But we were told that the hamiltonian density is ##H=T+V## which corresponds to the energy density (when you integrate it you get the actual hamiltonian which gives the energy for well behaved potentials). Did I miss-understand that?

You are misinterpreting what ##T## is in this case so you have the wrong Hamiltonian. I suggest you look at the full expression for the energy-momentum tensor and extract the time-time component, which is the energy density.

You are misinterpreting what ##T## is in this case so you have the wrong Hamiltonian. I suggest you look at the full expression for the energy-momentum tensor and extract the time-time component, which is the energy density.

Oh, right! So I got this for the energy density: $$\frac{1}{2}\dot{\phi}^2-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ Is it obvious that ##\phi## has to be a constant?

So the way I did it is to use $$T_{\mu\nu}=\frac{\partial L}{\partial(\partial_\mu\phi)}\partial_\nu\phi-g_{\mu\nu}L$$ which gives $$T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi-g_{\mu\nu}(\frac{1}{2}(\partial_\mu \phi)^2+\frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4)$$ $$T_{\mu\nu}=\frac{1}{2}\partial_\mu \phi\partial_\nu \phi-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ Now the energy density is given by $$T_{00}=\frac{1}{2}\partial_0 \phi\partial_0 \phi-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ Where is my mistake? Thank you!

Ah, ok. So it should be like this: $$T_{00}=\frac{1}{2}\partial_0\phi\partial_0\phi+\frac{1}{2}( \partial_x\phi\partial_x\phi+\partial_y\phi\partial_y\phi+\partial_z\phi\partial_z\phi)-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ I hope this is correct now. But still, why is this a minimum for constant ##\phi##? If the mass term would have been added and not subtracted, that would have made sense, but I am not sure I see it as obvious here? Thank you and sorry for taking so long!

The shape of the potential is irrelevant. What is relevant are the derivative terms that are always non-negative and zero only if the field is constant.

Hmm that makes it more clear. However, I am still a bit confused. The potential energy minimum is not necessarily attained for a constant value of ##\phi##, I mean ##\int V(x)##. So can't it happen that for some non-constant value of ##\phi## the kinetic term will indeed get bigger (as it is non-zero anymore) but the potential gets down by a higher amount and hence ##\int T+V## it's smaller overall? Intuitively makes sense that the more you move the more energy you have, but mathematically is not really obvious to me. Can I derive this using calculus of variation, something like ##\frac{\delta E}{\delta \phi}=0## and from here infer that ##\phi## is a constant?